7-3volumediskwasher.pptx

7.3 Volumes Disk and Washer Methods

2

Suppose I start with this curve.

y x

My boss at the ACME Rocket Company has assigned me to build a nose cone in this shape.

1

0

1

2

3

4

So I put a piece of wood in a lathe and turn it to a shape to match the curve.

2

How could we find the volume of the cone?

y x

One way would be to cut it into a series of thin slices (flat cylinders) and add their volumes.

1

0

1

2

3

4

The volume of each flat cylinder (disk) is:

 r 2  the thickness



 x

2

dx

In this case: r= the y value of the function thickness = a small change in x = dx

2

The volume of each flat cylinder (disk) is:

y x

 r 2  the thickness

1

 0

1

2

3

 x

2

dx

4

If we add the volumes, we get:

   x 4

0

2

dx

4

   x dx 0



 2

4

 8

x2 0

This application of the method of slicing is called the disk method. The shape of the slice is a disk, so we use the formula for the area of a circle to find the volume of the disk. If the shape is rotated about the x-axis, then the formula is: b

V    y 2 dx a

b

A shape rotated about the y-axis would be:

V    x 2 dy a

1 The region between the curve x  , 1  y  4 and the y

y-axis is revolved about the y-axis. Find the volume. y

x

1

1

2

3 4

1  .707 2 1  .577 3 1 2

We use a horizontal disk. The thickness is dy.

4

3

2

The radius is the x value of the 1 function  .

dy

y

1

2

 1  V   dy  y  1  

4

 

4

0

1

1

1 dy y

volume of disk

0

2   ln y 1    ln 4  ln1   ln 2  2 ln 2

4

y

The natural draft cooling tower shown at left is about 500 feet high and its shape can be approximated by the graph of this equation revolved about the y-axis:

500 ft

x

x  .000574 y 2  .439 y  185

The volume can be calculated using the disk method with a horizontal disk.



500

0

.000574 y

2

 .439 y  185 dy  24, 700, 000 ft 3 2

Example: Using calculus, derive the formula for finding the volume of a sphere of radius r. The region bounded by a semicircle and its diameter shown below is revolved about the x-axis, which gives us a sphere of radius r. Area of each cross section? (circle) 2 What is y?

A  y

A dx

–r

r

Equation of semicircle:

x y r 2

2

2

r 

2

x

A r  x 2

y r x 2

2

2

2





2

Example: Using calculus, derive the formula for finding the volume of a sphere of radius r. Area of each cross section? (circle)



A r  x 2

2



Volume: r

V    (r  x )dx 2

–r

r

r

2

(Remember r is just a number!!!) r 3

 2 1  V   r x  x  3  r   2 1 3   2 1 3  V    r r  r    r (r )  (r )  3   3  

Example: Using calculus, derive the formula for finding the volume of a sphere of radius r. Volume:

 2 1 3   2 1 3  V    r r  r    r (r )  (r )  3   3   –r

r

 3 1 3   3 1 3  V     r  r     r  r   3   3   2 3 2 3 4 3 V  r  r   r 3 3  3

Example: Find the volume of the solid generated when the region bounded by y = x2, x = 2, and y = 0 is rotated about the line x = 2. Area of each cross section? (circle)

A  r dy r

2

What is r?

A   2  x 

2

r=2–x

Remember, we are using a dy here!!! So, 2

2

yx

yx



A 2 y



2

Example: Find the volume of the solid generated when the region bounded by y = x2, x = 2, and y = 0 is rotated about the line x = 2. Area of each cross section? (circle)



A 2 y



2

Volume:

r

V    (2  y 2 )dy 1

2

Bounds? From 0 to intersection (y-value!!!)

8 V    (2  y )dy  3 0 4

1

2

4

3

y  2x

2

y  x2

The region bounded by y  x 2 and y  2 x is revolved about the y-axis. Find the volume.

1

If we use a horizontal slice: yx

y  2x y x 2

2

yx

0

1

2

The volume of the washer is:  V   0   4

 y

2

 y   2

2

  dy 

1   V     y  y 2  dy 0 4   4

V 

4

0

1 2 y  y dy 4

The “disk” now has a hole in it, making it a “washer”.

 R   r   thickness   R  r  dy 2

2

2

2

outer radius 4

1  1    y 2  y3  12  0 2

inner radius

 16    8   3 

8  3

This application of the method of slicing is called the washer method. The shape of the slice is a circle with a hole in it, so we subtract the area of the inner circle from the area of the outer circle.

The washer method formula is:

b

V    R2  r 2 dx a

y  x2

4

3

y  2x

2

1

0

1

r

y  2x y x 2

y  x2

yx

2

r  2 y y2    4  2 y   4  4 y  y dy 0 4 1 4 1 2    3 y  y  4 y 2 dy 0 4 4

V    R2  r 2 dy 0

2



y    2   2 y 0 2 

 dy 2





4

 3 2 1 3 8      y  y  y  12 3 0  2 16 64  8      24     3 3 3  3 2

 y2      4  2 y    4  4 y  y dy 0 4   4

The outer radius is: y R  2 2 The inner radius is:

R

4

4

If the same region is rotated about the line x=2: